# How do you deal with absolute values in a function when calculating volumes of solids of revolution?

For example, when calculating the volume of a solid of revolution for the area between the curves y=-abs(x-4)+4 ; y=0 about the x axis. I know the equation should be set up as pi * integral between 0 and 8 of (-abs(x-4)+4)^2 dx , but am unsure of how to proceed thereafter.

• Split the integral on $[0,8]$ into the sum of the integrals on $[0,4]$ and $[4,0]$. On the first $|x-4|=4-x$ on the second $|x-4|=x-4$. – arts Dec 1 '17 at 2:19
• I kind of get where you are going with this. Is this because r^2 can be thought of as r * r ? Why would each of the two integrals we split from the sum be 0 to 4 instead of 0 to 8? – foobarbaz Dec 1 '17 at 2:29
• That was a typo. The first integral is on $[0,4]$ and the second on $[4,8]$. – arts Dec 1 '17 at 2:31
• So do you just split the boundaries in half? i.e. if the original boundaries were 0 to 10 would I do [0,5] and then [5,10]? – foobarbaz Dec 1 '17 at 2:32
• No, split it at the $4$ because it is the point at which what is inside the absolute value ($x-4$) changes signs. – arts Dec 1 '17 at 2:33

I've done this volume calculation using Pappus's $(2^{nd})$ Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid R, i.e., $2πR$. The volume is simply $V=2\pi RA$.
The figure below shows the area to be revolved about the $x$-axis. The centroid of a triangle is straightforward and is shown here by the asterisk $x_c,y_c=4,4/3$ and the area is simply $A=16$.
$$V=2\pi\frac{4}{3}16=\frac{128\pi}{3}$$